- #1
Coop
- 40
- 0
Homework Statement
A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin60 = 0.87; cos60 = 0.5. Ignore friction and the weights of the pulleys.)
A) 50 J
B) 100 J
C) 174 J
D) 200 J
http://www.bestsamplequestions.com/mcat-sample-questions/mcat-sample-questions-physical-sciences/images/mcat-sample-questions-physical-science-ques-57.gif
Homework Equations
[tex]W=Fdcos(\theta)[/tex]
The Attempt at a Solution
I get that it takes 200 J of energy to raise a 4-kg weight 5 m. But I don't see how the force F does 200 J of work when referring to the relevant equation. Since theta is the angle between the force F and velocity of object, why would you not say [tex]W=(4 kg)(10\frac{m}{s^2})(5 m)cos(150)[/tex]? This would give 174 J. I guess what I am confused about is why is the theta said to be 0?
Thanks